Problem: The lifespans of lizards in a particular zoo are normally distributed. The average lizard lives $3.1$ years; the standard deviation is $0.7$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lizard living longer than $2.4$ years.
Explanation: $3.1$ $2.4$ $3.8$ $1.7$ $4.5$ $1$ $5.2$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $3.1$ years. We know the standard deviation is $0.7$ years, so one standard deviation below the mean is $2.4$ years and one standard deviation above the mean is $3.8$ years. Two standard deviations below the mean is $1.7$ years and two standard deviations above the mean is $4.5$ years. Three standard deviations below the mean is $1$ years and three standard deviations above the mean is $5.2$ years. We are interested in the probability of a lizard living longer than $2.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the lizards will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the lizards will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $2.4$ years and the other half $({16\%})$ will live longer than $3.8$ years. The probability of a particular lizard living longer than $2.4$ years is ${68\%} + {16\%}$, or $84\%$.